JEE Advanced 1992

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let a circle be given by $2 x (x - a) + y (2 y - b) = 0, (a \neq = 0, b \neq = 0)$ . Find the condition on $a$ and $b$ if two chords, each bisected by the $x$ -axis, can be drawn to the circle from $(a, b /2)$ .

Q: Wait, why do we use $T = S_1$? Can't we just find the slope and use the line equation?

You absolutely could, but $T = S_1$ is the 'secret weapon' of coordinate geometry. It bypasses the tedious slope-intercept calculations and directly relates the midpoint to the chord's equation. It's elegant, efficient, and less prone to calculation errors—exactly what you need in the heat of a JEE exam.

Q: Why do we need $D > 0$ for the discriminant? Why not $D \geq 0$?

The problem asks for 'two' chords. If $D = 0$, the quadratic has only one root, meaning there is only one unique chord. To have two distinct chords, we must have two distinct values for $h$, which strictly requires $D > 0$.

Visualized Solution

Visualizing the Circle and Point $P$

We are given the equation of a circle: $2 x (x - a) + y (2 y - b) = 0$ .
We also have a specific point $P (a, \frac{b}{2})$ .
Our goal is to find the condition on $a$ and $b$ such that we can draw exactly two chords from $P$ that are bisected by the $x$ -axis.

Converting to Standard Form

Let's expand the given equation: $2 x^{2} - 2 a x + 2 y^{2} - b y = 0$ .
To find the standard form, we divide the entire equation by $2$ .
This gives: $x^{2} + y^{2} - a x - \frac{b}{2} y = 0$ .

Defining the Midpoint $(h, 0)$

Since the chords are bisected by the $x$ -axis, their midpoints must lie on the $x$ -axis.
Let the midpoint of a chord be $M (h, 0)$ .
Any point on the $x$ -axis has a $y$ -coordinate of $0$ .

Using the Chord Formula $T = S_{1}$

The equation of a chord of a circle $S = 0$ with a given midpoint $(x_{1}, y_{1})$ is given by $T = S_{1}$ .
Here, the circle is $S = x^{2} + y^{2} - a x - \frac{b}{2} y = 0$ .
The midpoint is $(x_{1}, y_{1}) = (h, 0)$ .

Constructing $T$ and $S_{1}$

For $T$ , replace $x^{2} \to x h$ , $y^{2} \to y (0)$ , $x \to \frac{x + h}{2}$ , and $y \to \frac{y + 0}{2}$ .
$T = x h + 0 - \frac{a}{2} (x + h) - \frac{b}{4} (y + 0)$
For $S_{1}$ , substitute $(h, 0)$ into the circle equation:
$S_{1} = h^{2} + 0^{2} - ah - 0 = h^{2} - ah$

The Equation of the Chord

Equating $T = S_{1}$ :
$x h - \frac{a}{2} (x + h) - \frac{b}{4} y = h^{2} - ah$
This is the general equation of any chord of this circle bisected at $(h, 0)$ .

Passing through Point $P (a, \frac{b}{2})$

Since the chords are drawn from $P (a, \frac{b}{2})$ , this point must lie on the chord.
Substitute $x = a$ and $y = \frac{b}{2}$ into the chord equation:
$a (h) - \frac{a}{2} (a + h) - \frac{b}{4} (\frac{b}{2}) = h^{2} - ah$

Formulating the Quadratic in $h$

Expand the terms:
$ah - \frac{a ^{2}}{2} - \frac{ah}{2} - \frac{b ^{2}}{8} = h^{2} - ah$
Rearrange all terms to one side to form a standard quadratic $A h^{2} + B h + C = 0$ :
$h^{2} - \frac{3 a}{2} h + \frac{a ^{2}}{2} + \frac{b ^{2}}{8} = 0$

Condition for Two Distinct Chords

For two distinct chords to exist, there must be two distinct real values of $h$ .
This means the quadratic equation in $h$ must have two distinct real roots.
Therefore, the discriminant $D$ must be strictly greater than zero: $D > 0$ .
$D = B^{2} - 4 A C = (- \frac{3 a}{2})^{2} - 4 (1) (\frac{a ^{2}}{2} + \frac{b ^{2}}{8}) > 0$

Solving the Inequality

Expand the discriminant:
$\frac{9 a ^{2}}{4} - 2 a^{2} - \frac{b ^{2}}{2} > 0$
Simplify the $a^{2}$ terms: $\frac{a ^{2}}{4} - \frac{b ^{2}}{2} > 0$
Multiply by $4$ : $a^{2} - 2 b^{2} > 0 ⟹ a^{2} > 2 b^{2}$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given circle equation is

2 x (x - a) + y (2 y - b) = 0

. To simplify this, we expand the terms to get

2 x^{2} - 2 a x + 2 y^{2} - b y = 0

Dividing the entire equation by

2

, we obtain the standard form:

x^{2} + y^{2} - a x - \frac{b}{2} y = 0

This circle has its center at

(\frac{a}{2}, \frac{b}{4})

. This equation serves as our foundation for all subsequent geometric derivations.

The Midpoint Mystery

The problem specifies that the chords are bisected by the

x

-axis. Any point on the

x

-axis has a

y

-coordinate of

0

Let the midpoint of such a chord be

M (h, 0)

. Here,

h

is a variable parameter that defines the specific chord being bisected.

The Power of $T = S_{1}$

We utilize the theorem that the equation of a chord with a known midpoint

(x_{1}, y_{1})

is given by

T = S_{1}

. For our circle

S : x^{2} + y^{2} - a x - \frac{b}{2} y = 0

and midpoint

(h, 0)

, we calculate:

T = x h + y (0) - \frac{a}{2} (x + h) - \frac{b}{4} (y + 0)

S_{1} = h^{2} + 0^{2} - ah - 0 = h^{2} - ah

Equating

T = S_{1}

, we derive the general equation for any chord bisected at

(h, 0)

x h - \frac{a}{2} (x + h) - \frac{b}{4} y = h^{2} - ah

The Quadratic Climax

We are given that these chords must pass through the point

P (a, \frac{b}{2})

. Substituting these coordinates into our chord equation, we get:

a (h) - \frac{a}{2} (a + h) - \frac{b}{4} (\frac{b}{2}) = h^{2} - ah

Simplifying the expression leads to:

ah - \frac{a ^{2}}{2} - \frac{ah}{2} - \frac{b ^{2}}{8} = h^{2} - ah

Rearranging the terms into a standard quadratic equation in

h

, we find:

h^{2} - \frac{3 a}{2} h + \frac{a ^{2}}{2} + \frac{b ^{2}}{8} = 0

Final Calculation

For two distinct chords to exist, the quadratic equation must yield two distinct real values for

h

. This requires the discriminant

D

to be strictly greater than zero.

D = (- \frac{3 a}{2})^{2} - 4 (1) (\frac{a ^{2}}{2} + \frac{b ^{2}}{8}) > 0

Expanding this inequality:

\frac{9 a ^{2}}{4} - 2 a^{2} - \frac{b ^{2}}{2} > 0

\frac{a ^{2}}{4} - \frac{b ^{2}}{2} > 0

Multiplying by

4

, we arrive at the final condition:

$a^{2} > 2 b^{2}$

Common Traps & FAQs

Wait, why do we use

T = S_{1}

? Can't we just find the slope and use the line equation?

You absolutely could, but

T = S_{1}

is the 'secret weapon' of coordinate geometry. It bypasses the tedious slope-intercept calculations and directly relates the midpoint to the chord's equation. It's elegant, efficient, and less prone to calculation errors—exactly what you need in the heat of a JEE exam.

Why is the midpoint

(h, 0)

? Shouldn't it be

(h, k)

Great catch! The problem explicitly states the chords are bisected by the

x

-axis. By definition, any point on the

x

-axis has a

y

-coordinate of zero. So, while the midpoint is generally

(h, k)

, here

k

is forced to be

0

. This constraint is our key to solving the problem.

Why do we need

D > 0

for the discriminant? Why not

D \geq 0

The problem asks for 'two' chords. If

D = 0

, the quadratic has only one root, meaning there is only one unique chord. To have two distinct chords, we must have two distinct values for

h

, which strictly requires

D > 0

Animated Solution for Mathematics - Circles: Let a circle be given by $2 x (x - a) + y (2 y - b) = 0, (a \neq = 0, b \neq = 0)$ . Find the condition on $a$ and $b$ if two chords, each bisected by the $x$ -axis, can be drawn to the circle from $(a, b /2)$ .

Visualized Solution

Visualizing the Circle and Point $P$

Converting to Standard Form

Defining the Midpoint $(h, 0)$

Using the Chord Formula $T = S_{1}$

Constructing $T$ and $S_{1}$

The Equation of the Chord

Passing through Point $P (a, \frac{b}{2})$

Formulating the Quadratic in $h$

Condition for Two Distinct Chords

Solving the Inequality

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Midpoint Mystery

The Power of $T = S_{1}$

The Quadratic Climax

Final Calculation

Similar Questions

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List-I

List-II

Visualized Solution

Visualizing the Circle and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Converting to Standard Form

Defining the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (h,0)

Using the Chord Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=S1​

Constructing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

The Equation of the Chord

Passing through Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(a,2b​)

Formulating the Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Condition for Two Distinct Chords

Solving the Inequality

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Midpoint Mystery

The Power of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=S1​

The Quadratic Climax

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } From the origin chords are drawn to the circle (x−1)2+y2=1. The equation of the locus of the mid-points of these chords is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the circle C1​:x2+y2−2x−1=0 at the point (2, 1) cuts off a chord of length 4 from a circle C2​ whose centre is (3, -2). The radius of C2​ is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

List-I

List-II

Visualizing the Circle and Point $P$

Defining the Midpoint $(h, 0)$

Using the Chord Formula $T = S_{1}$

Constructing $T$ and $S_{1}$

Passing through Point $P (a, \frac{b}{2})$

Formulating the Quadratic in $h$

The Power of $T = S_{1}$

From the origin chords are drawn to the circle $(x - 1)^{2} + y^{2} = 1$ . The equation of the locus of the mid-points of these chords is .........

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ is :-

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is